poj 3233 Matrix Power Series(midhard)

http://poj.org/problem?id=3233

　　矩阵快速幂的题，求ΣA^i(1≤i≤k)的矩阵并输出。将它二分，用快速幂的方法来O(nlogn)的复杂度完成计算。

代码如下：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;
typedef int ll;
const int matSize = 31;
const int stdMod = 1000000007;
int calSize = matSize;
int mod = stdMod;

struct Matrix {
    ll val[matSize][matSize];

    Matrix(bool Init = false) {
        for (int i = 0; i < calSize; i++) {
            for (int j = 0; j < calSize; j++) {
                val[i][j] = 0;
            }
            if (Init) val[i][i] = 1;
        }
    }

    void print() {
        for (int i = 0; i < calSize; i++) {
            for (int j = 0; j < calSize; j++) {
                if (j) putchar(' ');
                printf("%d", val[i][j]);
            }
            puts("");
        }
    }
} Base;

Matrix operator * (Matrix &_a, Matrix &_b) {
    Matrix ret = Matrix();

    for (int i = 0; i < calSize; i++) {
        for (int k = 0; k < calSize; k++) {
            if (_a.val[i][k]) {
                for (int j = 0; j < calSize; j++) {
                    ret.val[i][j] += _a.val[i][k] * _b.val[k][j];
                    ret.val[i][j] %= mod;
                }
            }
        }
    }

    return ret;
}

Matrix operator ^ (Matrix &_a, ll _p) {
    Matrix ret = Matrix(true);
    Matrix __a = _a;

    while (_p) {
        if (_p & 1) {
            ret = ret * __a;
        }
        __a = __a * __a;
        _p >>= 1;
    }

    return ret;
}

Matrix operator + (Matrix &_a, Matrix &_b) {
    Matrix ret;

    for (int i = 0; i < calSize; i++) {
        for (int j = 0; j < calSize; j++) {
            ret.val[i][j] = (_a.val[i][j] + _b.val[i][j]) % mod;
        }
    }
    //puts("~~~~~~");
    //ret.print();

    return ret;
}

void deal(int k) {
    Matrix tmp = Base, ans = Matrix(), ep = Base, one = Matrix(true), as;

    while (k) {
//        puts("tmp");
//        tmp.print();
        if (k & 1) {
            ans = ans * ep;
            ans = ans + tmp;
        }
        as = one + ep;
        tmp = tmp * as;
        ep = ep * ep;
        k >>= 1;
//        puts("!!!");
//        ans.print();
//        puts("~~~");
    }

    ans.print();
}


int main() {
    int k;

//    freopen("in", "r", stdin);
    while (~scanf("%d%d%d", &calSize, &k, &mod)) {
        for (int i = 0; i < calSize; i++) {
            for (int j = 0; j < calSize; j++) {
                scanf("%d", &Base.val[i][j]);
                Base.val[i][j] %= mod;
            }
        }
        deal(k);
    }

    return 0;
}

——written by Lyon